CHM 8304
Kinetic analyses 1
CHM 8304 Physical Organic Chemistry
Kinetic analyses
Scientific method
• proposal of a hypothesis • conduct experiments to test this hypothesis
– confirmation – refutation
• the trait of refutability is what distinguishes a good scientific hypothesis from a pseudo-hypothesis (see Karl Popper)
2
CHM 8304
Kinetic analyses 2
“Proof” of a mechanism
• a mechanism can never really be ‘proven’ – inter alia, it is not directly observable!
• one can propose a hypothetical mechanism • one can conduct experiments designed to refute certain hypotheses • one can retain mechanisms that are not refuted • a mechanism can thus become “generally accepted”
3
Mechanistic studies
• allow a better comprehension of a reaction, its scope and its utility
• require kinetic experiments – rate of disappearance of reactants – rate of appearance of products
• kinetic studies are therefore one of the most important disciplines in experimental chemistry
4
CHM 8304
Kinetic analyses 3
Outline: Energy surfaces
• based on section 7.1 of A&D – energy surfaces – energy profiles – the nature of a transition state – rates and rate constants – reaction order and rate laws
5
Energy profiles and surfaces
• tools for the visualisation of the change of energy as a function of chemical transformations – profiles:
• two dimensions • often, energy as a function of ‘reaction coordinate’
– surfaces: • three dimensions • energy as a function of two reaction coordinates
6
CHM 8304
Kinetic analyses 4
Example: SN2 energy profile
• one step passing by a single transition state Fr
ee e
nerg
y
Reaction co-ordinate
X:- + Y-Z reactants
X---Y---Z transition state
δ- δ-
X-Y + :Z-
products
ΔG°
ΔG‡ activation energy
7
Reminder: activated complex • ethereal complex formed at the transition state of the reaction
– partially formed bonds – partially broken bonds
+ ClC + ClClCHOδ−δ−
HOH
HH HH
H
HO CH
HH
transition state not isolable partial bonds
very unstable; <10-13 s
8
CHM 8304
Kinetic analyses 5
Free energy profiles vs surfaces
• a reaction profile gives the impression that only one reaction pathway can lead to the formation of products
• however, in reality, reactant molecules are free to ‘explore’ all available reaction pathways – the pathway with the lowest energy is the one that determines the
predominant mechanism
• free energy surfaces are in 3D, and take account of more than one event that must take place during a reaction (e.g. bond cleavage and formation) – allow the consideration off ‘off-pathway’ interactions – allow the visualisation of mechanistic promiscuity
9
Energy surfaces
• allow the visualisation of several reaction pathways, or even several reactions that may take place
chem.wayne.edu reactant
product B
product A
activated complex for A activated complex for B
10
CHM 8304
Kinetic analyses 6
More-O’Ferrall and Jencks
• Rory More-O’Ferrall – University College, Dublin – mechanisms et catalysis
• William P. Jencks (1927-2007) – Biochemistry, Brandeis University – a father of modern physical organic chemistry
and biological chemistry – author of classic textbook – laureate of several awards
11
Profile to surface
X:- + Y-Z reactants
X-Y + :Z-
products
Cleavage of Y-Z bond
Form
atio
n of
X-Y
bon
d
X:- + Y+ + :Z- unstable
intermediates
X-Y--Z unstable
intermediate
‡
Free
ene
rgy
Reaction co-ordinate
X:- + Y-Z reactants
X---Y---Z TS
δ- δ-
X-Y + :Z-
products
2D: rxn. coord. & energy
2D: 2 rxn. coord.
3D: 2 rxn. coord. & energy
12
CHM 8304
Kinetic analyses 7
Mountain terminology
13
summit (Helmet Pk.) summit (Mt. Robson)
Helmet/Robson col
‘Summit’ vs ‘col’
• the predominant mechanism follows the reaction pathway of lowest activation energy
• the ‘summit’ (TS) of this pathway is often a ‘col’ between peaks of even higher energy
Free
ene
rgy
Rxn. co-ord. 14
CHM 8304
Kinetic analyses 8
Exercise: Diagram to mechanism • Draw the reaction profile that corresponds to the lowest energy
reaction pathway on the MOFJ diagram below. Draw the corresponding mechanism including its transition state(s).
Cleavage of C-LG bond
Form
atio
n of
C-Nu
c bon
d
+ Nuc+ GPEt LG
Nuc + GPLG
H3C CH2
Nuc
GPLG
GP + NucLG
Free
ene
rgy
Rxn. Co-ord.
15
R
R
R
R-
Multi-step reactions • when a reaction takes place via several elementary chemical steps, one of
these steps will limit the global rate of transformation – named the rate determining step (rds) or rate limiting step (rls)
• the rate limiting step is always the step having the highest energy transition state – even if this step does not have the largest microscopic activation barrier!
16
CHM 8304
Kinetic analyses 9
Rate-limiting step on a reaction profile
• the rate-limiting step is the one with the highest energy transition state with respect to the ground state
Free
ene
rgy
Rxn. Co-ord.
reactants
products
TS1
int.
TS2
ΔG‡
Free
ene
rgy
Rxn. Co-ord.
reactants
products
TS1
int.
TS2 activation energy of global reaction; step 2 is rls
ΔG1‡
ΔG2‡
ΔG1‡ > ΔG2
‡ so step 1 is rls? NO!!
17
Rate-limiting step on a reaction profile
• the rate-limiting step is the one with the highest energy transition state with respect to the ground state – see A&D, Figure 7.3
Free
ene
rgy
Rxn. co-ord.
ΔG‡
Free
ene
rgy
Rxn. co-ord.
ΔG‡
Free
ene
rgy
Rxn. co-ord.
ΔG‡
Free
ene
rgy
Rxn. co-ord.
ΔG‡
Free
ene
rgy
Rxn. co-ord.
ΔG‡
Free
ene
rgy
Rxn. co-ord.
ΔG‡
18
CHM 8304
Kinetic analyses 10
Rate vs rate constant
• the reaction rate depends on the activation barrier of the global reaction and the concentration of reactants, according to rate law for the reaction – e.g. v = k[A]
• the proportionality constant, k, is called the rate constant
19
Rate vs rate constant • reaction rate at a given moment is the instantaneous slope of [P] vs time • a rate constant derives from the integration of the rate law
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20
Temps
[Produit]
different initial rates
same half-lives; same rate constants
different concentrations of reactants; different end points
20
CHM 8304
Kinetic analyses 11
Rate law and molecularity • each reactant may or may not affect the reaction rate, according to the
rate law for a given reaction • a rate law is an empirical observation of the variation of reaction rate as
a function of the concentration of each reactant – procedure for determining a rate law:
• measure the initial rate (<10% conversion) • vary the concentration of each reactant, one after the other • determine the order of the variation of rate as a function of the concentration of
each reactant • e.g. v ∝ [A][B]2
• the order of each reactant in the rate law indicates the stoichiometry of its involvement in the transition state of the rate-determining step
21
Integers in rate laws
• integers indicate the number of equivalents of each reactant that are found in the activated complex at the rae-limiting transition state – e.g.:
• reaction: A + B à P – mechanism: A combines with B to form P
• rate law: v ∝ [A][B] – one equivalent of each of A and B are present at the TS of the rds
22
CHM 8304
Kinetic analyses 12
Fractions in rate laws
• fractions signify the dissociation of a complex of reactants, leading up to the rds: – e.g. :
• elementary reactions: A à B + B; B + C à P – mechanism: reactant A exists in the form of a dimer that must dissociate before
reacting with C to form P
• rate law: v ∝ [A]½[C] – true rate law is v ∝ [B][C], but B comes from the dissociation of dimer A – observed rate law, written in terms of reactants A and C, reflects the dissociation of A – it is therefore very important to know the nature of reactants in solution!
23
• negative integers indicate the presence of an equilibrium that provides a reactive species: – e.g. :
• elementary reactions: A B + C; B + D à P – mechanism: A dissociates to give B and C, before B reacts with D to give P
• rate law: v ∝ [A][C]-1[D] – true rate law is v ∝ [B][D], but B comes from the dissociation of A – observed rate law, written in terms of reactants A and D, reflects the dissociation of A – apparent inhibition by C reflects the displacement of the initial equilibrium
Negative integers in rate laws
24
CHM 8304
Kinetic analyses 13
Outline: Transition State Theory
• based on section 7.2 of A&D – Arrhenius approach – Eyring approach – temperature effects – interpretation of thermodynamic parameters
25
Arrhenius • Svante Arrhenius (1859-1927)
– Swedish physicist and chemist (Stockholm) – solution chemistry, activation energy – also: panspermia, universal language,
greenhouse effect, racial biology – one of the founders of the Nobel Institute – Nobel Prize (1903) for the electrolytic theory of
of dissociation
26
CHM 8304
Kinetic analyses 14
Arrhenius equation • in 1889, Arrhenius noted that the rate constant of a given reaction
increases exponentially with temperature :
• the activation energy, Ea, was defined as the energy necessary to convert reactants into “high energy species”
• however, this energy is not expressed in terms of thermodynamic parameters – for this, it was necessary to develop the transition state theory
2a
RTE
Tln
=dkd
AlnRTEln a +−=k
27
Arrhenius plot
1 / T (K-1)
ln k
obs pente = - Ea / R
ln A
Graphique d'Arrhenius
AlnRTEln a +−=k
28
slope
CHM 8304
Kinetic analyses 15
Eyring • Henry Eyring (Sr.) (1901-1981)
– American professor of theoretical chemistry and reaction rates (Princeton, Utah)
– laureate of several prizes: Wolf Prize, Priestley Medal et National Medal of Science
29
Transition State Theory (TST)
• developed by Eyring to explain observed reaction rates in terms of standard thermodynamic parameters
• in their transformation into products, reactants must attain a transition state, in the form of an activated complex – the reaction rate is proportional to the concentration of this activated complex
K‡ k‡A•B PA B+
complexeactivé
[A][B]]B•A[ ‡
‡ =K [A][B]]B•A[ ‡‡ K= [A][B]]B•A[ ‡‡‡‡ Kkkv ==
30
quasi equilibrium
activated complex
CHM 8304
Kinetic analyses 16
Eyring equation
• based on transition state theory, relating kinetic parameters to thermodynamic parameters – k‡ equals the vibration frequency, ν, that leads to decomposition of the
activated complex :
• kB is Boltzmann’s constant (kB = 1.3806 × 10-23 J K-1)
• h is Planck’s constant (h = 6.626 × 10-34 J�s)
– ΔG = -RT ln Keq so ΔG‡ = -RT ln K‡ and
K‡ k‡A•B PA B+
complexeactivé
[A][B]]B•A[ ‡‡‡‡ Kkkv ==
⎟⎟⎠
⎞⎜⎜⎝
⎛ −
= RTΔG
‡
‡
eK
⎟⎟⎠
⎞⎜⎜⎝
⎛ −
== RTGΔ
B‡‡obs
‡
T ehkKkk
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−
= RSΔ
RTHΔ
Bobs
‡‡
T ehkk
RSΔ
T1
RHΔ
Tln
‡‡
B
obs +⋅−
=⎟⎟⎠
⎞⎜⎜⎝
⎛
khk
hkk TB‡ ==
31
activated complex
Eyring plot
1 / T (K-1)
ln (k
obsh
/ k BT
)
pente = - ΔH‡ / R
Graphique d'Eyring
ΔS‡ / R
RSΔ
T1
RHΔ
Tln
‡‡
B
obs +⋅−
=⎟⎟⎠
⎞⎜⎜⎝
⎛
khk
32
slope =
CHM 8304
Kinetic analyses 17
Activation (free) energy, ΔG‡
• directly related to reaction rate • this parameter represents the sum of energetic factors that influence
reaction rate
33
Activation enthalpy, ΔH‡
• a measure of the energy barrier that must be overcome by the reactants • it is the amount of heat energy that the reactants must acquire during their
transformation into the activated complex – principally related to the formation and cleavage of bonds
34
CHM 8304
Kinetic analyses 18
Activation entropy, ΔS‡
• a measure of the fraction of all reactants having the necessary activation enthalpy to react that do, in reality, react
• a measure of probability, this parameter takes account of concentration and solvation effects, steric hindrance and functional group orientation – for example for a unimolecular reaction, ΔS‡ ≈ 0 – for multimolecular reactions, ΔS‡ < 0
• typically –5 to –40 cal mol-1 K-1 (“entropy units”, e.u.)
35
Exercise A: Eyring plot to mechanism • use the Eyring plot to determine the enthalpy and entropy of activation
for the substitution reaction of iPr-Br – are the data more consistent with SN1 or SN2 ?
Temperature dependence
-25.00
-20.00
-15.00
-10.00
-5.00
0.00
5.00
10.00
15.00
20.00
0.0E+00 5.0E-04 1.0E-03 1.5E-03 2.0E-03 2.5E-03 3.0E-03 3.5E-03 4.0E-03
1/T (K-1)
ln (k
obs h
/ k B
T)
R = 1.987 cal/mol/K
intercept = ΔS‡/R; ΔS‡ = -6 cal/mol/K
slope = -ΔH‡/R; ΔH‡ = 10 kcal/mol
more ordered TS
little change in bond order at TS
H
CH3H3C
BrNuδ− δ−
‡
SN2 36
CHM 8304
Kinetic analyses 19
Temperature dependence
-50.00
-40.00
-30.00
-20.00
-10.00
0.00
10.00
20.00
0.0E+00 5.0E-04 1.0E-03 1.5E-03 2.0E-03 2.5E-03 3.0E-03 3.5E-03 4.0E-03
1/T (K-1)
ln (k
obs h
/ k B
T)
Exercise B: Eyring plot to mechanism
R = 1.987 cal/mol/K
intercept = ΔS‡/R; ΔS‡ = 10 cal/mol/K
slope = -ΔH‡/R; ΔH‡ = 25 kcal/mol less ordered TS
net bond cleavage
at TS
SN1
H
CH3H3C
Brδ+ δ−
‡
37
• use the Eyring plot to determine the enthalpy and entropy of activation for the substitution reaction of iPr-Br – are the data more consistent with SN1 or SN2 ?
Effect of temperature • an increase of 10 °C will double the rate of most reactions
– due to the increase of the average energy of reactant molecules :
% o
f mol
cule
s hav
ing
a gi
ven
ener
gy
Energy
T1
T2
T2 > T1
ΔG‡
38
CHM 8304
Kinetic analyses 20
Rate constants and temperature
• by analogy with ΔG° = -RT ln Keq, ΔG‡ = -RT ln(k/k0)
or
• the more the temperature increases, the more the factor (ΔG‡/RT) decreases, and the more k increases
RTekk‡ΔG
0
−=
39
Outline: Postulates et principles
• based on section 7.3 of A&D – Hammond postulate – reactivity vs selectivity – Curtin-Hammett principle – microscopic reversibility – kinetic control vs thermodynamic control
40
CHM 8304
Kinetic analyses 21
Hammond • George S. Hammond (1921-2005)
– American chemistry professor – studied the relation between kinetics and product
distribution – laureate of Norris Award and Priestley Medal
41
Hammond Postulate • the structure of the activated complex at the transition state of an
elementary reaction that is…
…endergonique (ΔG° > 0) resembles the products
…exergonique (ΔG° < 0) resembles the reactants
Free
ene
rgy
Reaction co-ordinate
42
CHM 8304
Kinetic analyses 22
Hammond Postulate • to put it another way, the structure of the activated complex at the
transition step of an elementary reaction…
…varies according to ΔG° :
Free
ene
rgy
Reaction co-ordinate 43
Reactivity vs selectivity • in a comparison of two different possible reactions, the selectivity
between these two reactions may be affected by the reactivity of the reactant: – a very reactive molecule will pass by two TS having very similar structures,
without demonstrating great selectivity between them:
Free
ene
rgy
Reaction co-ordinate
reactant
product 2 product 1
small ΔΔG‡
44
CHM 8304
Kinetic analyses 23
• in a comparison of two different possible reactions, the selectivity between these two reactions may be affected by the reactivity of the reactant: – a relatively unreactive molecule will pass by two TS having very different
structures, showing great selectivity between them:
Reactivity vs selectivity Fr
ee e
nerg
y
Reaction co-ordinate
reactant
product 2
product 1
large ΔΔG‡
45
Curtin • David Y. Curtin (~1920 - )
– American chemist (UIUC) – studied the difference in reaction rate between
different isomers
46
CHM 8304
Kinetic analyses 24
Hammett • Louis Hammett (1894-1987)
– American physical chemist (Columbia) – studied the correlation of structure and function – credited with inventing the expression “physical
organic chemistry” – laureate of awards from the National Academy of
Science, two Norris Awards, Priestley Medal
47
Curtin-Hammett Principle
• for most organic compounds, conformational changes are more rapid than chemical transformations
• consider the case where two conformers react to give two different products :
– ... and where the conformational equilibrium is faster than the reactions that lead to the formation of these products :
A' A"kykx
X Y
k1
k1
k1 >> kx et k -1 >> ky
48
and
CHM 8304
Kinetic analyses 25
Curtin-Hammett Principle
• a priori, one cannot predict the proportion of products X et Y, based exclusively on the relative stabilities of A' and A"
• the following energy diagram illustrates that the the relative rates of formation of X and Y depend rather on the relative energies of the transition states leading to their formation
49
Curtin-Hammett Principle
X
Y
A' A"
Rxn. coord.
Free
ene
rgy
K = [A"]/[A'] = k1/k-1 = e-ΔG°/RT
ΔG°
ΔG‡y
ΔG‡x
ΔΔG‡
d[X]/dt = kx[A']
d[Y]/dt = ky[A"]
d[Y]/dt d[X]/dt
= (ky/kx) K
kx[A'] ky[A"]
=
‡ y e-ΔG /RT
= ‡ x e-ΔG /RT
× e-ΔG°/RT = e(ΔG -ΔG -ΔG°)/RT ‡ y
‡ x
= eΔΔG /RT ‡
50
CHM 8304
Kinetic analyses 26
Curtin-Hammett: Example #1 • methylation of a tertiary bicyclic amine:
– reactant conformational equilibrium does not determine product distribution
NMe
NMeH3C
13CH3IN
MeN
Me CH313CH3I
more stable
less stable
fast slow major minor
Rxn. co-ordinate
Free
ene
rgy
ΔΔG‡
51
Curtin-Hammett Conditions • it is important to keep the following condition in mind: conformational
changes must be faster than chemical transformations – in this case, [PA]/[PB] = kA/kB×Keq
• however, at low temperatures, conformational changes are slow, and product ratios often reflect the composition of the initial equilibrium (“kinetic quench”) – i.e. kx >> k1 and ky >> k-1, so [PA]/[PB] = Keq
– without interconversion, A and B are essentially separate populations that react independently
• all A is converted to PA and all B is converted to PB, regardless of the relative rates of these transformations
k1 >> kx et k -1 >> ky
52
and
CHM 8304
Kinetic analyses 27
N
O
Me
Me
CH3
CH3-BrN
O
Me
Me
CH3
CH3-Br
fast fast N
NMe
H
Me
OOMe
Me
H
N
O
Me
Me
s-BuLi
ΔG = -3.0 kcalby calculations
“Kinetic quench”: Example • alkylation of pyrrolidinone at -78°C
JACS, 1997, 119, 4565
disfavoured
slow, at -78°C
Rxn. co-ordinate
Free
ene
rgy
ΔG‡A
99% 1%
ΔG‡B
[ ][ ]
[ ][ ]0
0
A
B
AB
PP
=
53
Curtin-Hammett: Conclusions
• Curtin-Hammett principle is very important to keep in mind: – a priori, one cannot correlate reactant equilibrium with product distribution
• for fast conformational equilibria and relatively slow reactions, product ratio is determined by ΔΔG‡
• however, for slow conformational equilibria and relatively fast reactions, product ratio is determined by Keq
• in practice, it is difficult to apply this principle : • it requires the measurement of the proportion of reactant conformers • it requires independent measurement of equilibrium conformation and reaction
rates
54
CHM 8304
Kinetic analyses 28
Microscopic reversibility
• the reaction pathway for a reverse reaction must be exactly the reverse of the reaction pathway for the forward reaction – the transition states for the forward and reverse reactions are identical – allows the prediction of the nature of a transition state for one reaction, based
on knowledge of the transition state for the reverse reaction :
general base catalysis
general acid catalysis
55
H3C OH
O
OEtH
H3C OH
O
OEt
BB
H
Microscopic reversibility
• the reaction pathway for a reverse reaction must be exactly the reverse of the reaction pathway for the forward reaction – the transition states for the forward and reverse reactions are identical – allows the prediction of the nature of a transition state for one reaction, based
on knowledge of the transition state for the reverse reaction :
56
H3C OH
O
O
H3C OH
O
OO2N O2N
uncatalysed attack
uncatalysed expulsion
CHM 8304
Kinetic analyses 29
Kinetic vs thermodynamic control • factors that can influence product ratios • kinetic control:
– ratio affected by ΔΔG‡, not by ΔΔGrxn – irreversible reactions, Curtin-Hammett conditions
Reaction coordinate
Free
ene
rgy
ΔΔG‡
reactant product B product A
57
Kinetic vs thermodynamic control • factors that can influence product ratios • thermodynamic control:
– ratio affected by ΔΔGrxn not by ΔΔG‡
– reversible reactions (often at high temperatures)
58
Reaction coordinate
Free
ene
rgy
ΔΔGrxn
reactant product B product A
CHM 8304
Kinetic analyses 30
Outline: Experimental kinetics
• based on A&D section 7.4 – practical kinetics – kinetic analyses
• first order • second order • pseudo-first order
59
Practical kinetics
1. development of a method of detection (analytical chemistry!) 2. measurement of concentration of a product or of a reactant as a function
of time 3. measurement of reaction rate (slope of conc/time; d[P]/dt or -d[A]/dt)
– correlation with rate law and reaction order 4. calculation of rate constant
– correlation with structure-function studies
60
CHM 8304
Kinetic analyses 31
Kinetic assays
• method used to measure the concentration of reactants or of products, as a function of time – often involves the synthesis of chromogenic or fluorogenic reactants
• can be continuous or non-continuous
61
Continuous reaction assay
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20
Time
[Pro
duct
]
Continuous assay • instantaneous detection of reactants or products as the reaction is
underway – requires sensitive and rapid detection method
• e.g.: UV/vis, fluorescence, IR, (NMR), calorimetry
Continuous reaction assay
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20
Time
[Pro
duct
]
62
CHM 8304
Kinetic analyses 32
Continuous reaction assay
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20
Time[P
rodu
ct]
Discontinuous assay • involves taking aliquots of the reaction mixture at various time points,
quenching the reaction in those aliquots and measuring the concentration of reactants/products – wide variety of detection methods applicable
• e.g.: as above, plus HPLC, MS, etc.
63
Continuous reaction assay
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20
Time[P
rodu
ct]
Discontinuous
Initial rates • the first ~10 % of a reaction is almost linear, regardless of the
order of a reaction • ΔC vs Δt gives the rate, but the rate constant depends on the
order of the reaction
0 10 20 30 40 50 60 70 80 90 100
Temps (min)
0
10
20
30
40
50
60
70
80
90
100
[A] (
mM
)
k = 5 mM min- 1 k1 = 0.07 min- 1
k2 = 0.001 mM- 1 min- 1
tΩ = 10 min
Time
64
½
CHM 8304
Kinetic analyses 33
Calculation of a rate constant
• now, how can a rate constant, k, be determined quantitatively? • the mathematical equation to use to determine the value of k differs
according to the order of the reaction in question – the equation must be derived from a kinetic scheme – next, the data can be “fitted” to the resulting equation using a computer
(linear, or more likely, non-linear regression)
65
The language of Nature • “Nul ne saurait comprendre la nature si
celui-ci ne connaît son langage qui est le langage mathématique”
- Blaise Pascal – (‘None can understand Nature if one does not know its
language, which is the language of mathematics’)
• Natural order is revealed through special mathematical relationships
• mathematics are our attempt to understand Nature – exponential increase: the value of e – volume of spherical forms: the value of π
Blaise Pascal, French mathematician
and philosopher, 1623-1662
66
CHM 8304
Kinetic analyses 34
First order (simple)
A Pk1
= k1 [A]Vitesse = v = = -d[P]dt
d[A]dt
d[P] k ([P] [P])1dt= −∞
d[P] k ([A] [P])1 0dt= −
d dt[P][P] [P]
k1∞ −
=
∫∫ =−∞
t
01
P
0dtk
[P][P][P]d
ln ([P]∞ - [P]0) - ln ([P]∞ - [P]) = k1 t
ln t[A][A]
k01=
[A] = [A]0k1e− t
linear relation
mono-exponential decrease
ln t[A]½[A]
k0
01 ½=
t ln½
1
2k
=half-life
Rate
67
First order (simple)
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
kobs = k1
[A]∞
[A]0
tΩ
0 10 20 30 40 50 60 70 80 90 100Temps
-7-6-5-4-3-2-101
ln ([
A]/[A
] 0)
pente = kobs = k1
mono-exponential decrease
Time
Time
slope
68
t½
CHM 8304
Kinetic analyses 35
First order (reversible)
A Pk1
k-1
= k1 [A] - k-1 [P]Vitesse = v = = -d[P]dt
d[A]dt
( )( )[P]kk[P]k[A]k
[P]k[P][P][A]kdt[P]d
dt[A]d
1-1éq1éq1
1-éqéq1
+−+=
−−+==−
À l’équilibre, k1 [A]éq = k-1 [P]éq ( )
( )( )− = = + − +
= + −
ddt
ddt
[A] [P]k [P] k [P] k k [P]
k k [P] [P]
-1 éq 1 éq 1 -1
1 -1 éq
( )ddtt[P]
[P] [P]k k
éq0
P
1 -1 0−= +∫ ∫
( )( ) ( )ln t[P] [P]
[P] [P]k k
éq 0
éq
1 -1
−
−= +linear relation kobs = (k1 + k-1)
Rate
At equilibrium,
69
First order (reversible)
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
[A]Èq
[P]Èq
kobs = k1
kobs = k1 + k-1
[A]∞
[A]0
[P]0
( )t
ln½
1
2k k
=+ −1
Kéq 1
-1
kk
=
Time
70
CHM 8304
Kinetic analyses 36
Second order (simple)
Vitesse = v = = k2[A][B]d[P]dt
A B P+k2
d[P] k ([A] [P])([B] [P])2 0 0dt= − −
d dt[P]([A] [P])([B] [P])
k0 0
2− −=
ddtt[P]
([A] [P])([B] [P])k
0 00
P
2 0− −=∫ ∫
1[B] [A]
[A] ([B] [P])[B] ([A] [P])
k0 0
0 0
0 02−
−
−
⎛
⎝⎜
⎞
⎠⎟ =ln t
a lot of error is introduced when [B]0 and [A]0 are similar
Rate
71
Second order (simplified)
If [A]0 = [B]0 :
Vitesse = v = = k2[A][B]d[P]dt
A B P+k2
ddtt[P]
([A] [P])k
020
P
2 0−=∫ ∫
1 1[A] [P] [A]
k0 0
2−− = t
1 1[A] [A]
k0
2− = tlinear relation
t½2 0
1k [A]
=half-life
1 1[A]2
[A]k
0 02 ½⎛
⎝⎜
⎞⎠⎟
− = t
Rate
72
CHM 8304
Kinetic analyses 37
Second vs first order
• one must often follow the progress of a reaction for several (3-5) half-lives, in order to be able to distinguish between a first order reaction and a second order reaction :
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
premier ordre, k1
k2 = k1 ˜ 70
k2 = k1
first order,
73
Réactions cinétiques
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20 25
Temps
[Pro
duit]
Calcul de constante de vitesse
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
0 20 40 60 80 100 120
[A]
vite
sse
initi
ale
Example: first or second order? • measure of [P] as a function of time • measure of v0 as a function of time [A]
Reaction kinetics
first order [A] = [A]0 e –k1t
Time
[Pro
duct
]
Rate constant calculation
v0 = k1 [A]
In
itial
rate
74
CHM 8304
Kinetic analyses 38
Réactions cinétiques
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20 25
Temps
[Pro
duit]
Calcul de constante de vitesse
0.00
5.00
10.00
15.00
20.00
25.00
30.00
0 20 40 60 80 100 120
[A] = [B]
vite
sse
initi
ale
second order
• measure of [P] as a function of time • measure of v0 as a function of time [A]
Time
Reaction kinetics
[Pro
duct
]
Rate constant calculation
In
itial
rate
1/[A] – 1/[A]0 = k2t v0 = k2 [A][B]
Example: first or second order?
75
Pseudo-first order
• in the case where the initial concentration of one of the reactants is much larger than that of the other, one can simplify the treatment of the experimental data
Vitesse = v = = k2[A][B]d[P]dt
A B P+k2
1[B] [A]
[A] ([B] [P])[B] ([A] [P])
k0 0
0 0
0 02−
−
−
⎛
⎝⎜
⎞
⎠⎟ =ln t general solution:
second order:
76
CHM 8304
Kinetic analyses 39
Pseudo-first order • consider the case where [B]0 >> [A]0 (>10 times larger)
– the concentration of B will not change much over the course of the reaction; [B] = [B]0 (quasi-constant)
1[B]
[A] [B][B] ([A] [P])
k0
0 0
0 02ln
−
⎛
⎝⎜
⎞
⎠⎟ = t
1[B] [A]
[A] ([B] [P])[B] ([A] [P])
k0 0
0 0
0 02−
−
−
⎛
⎝⎜
⎞
⎠⎟ =ln t
ln [A]([A] [P])
k [B]0
02 0−
= t
ln [A]([A] [P])
k où k k [B]0
01'
1'
2 0−= =t
ln t[A][A]
k01'=
[A] = [A]0k1'
e− tmono-exponential decrease
where
77
Pseudo-first order • from a practical point of view, it is more reliable to determine a
second order rate constant by measuring a series of first order rate constants as a function of the concentration of B (provided that [B]0 >> [A]0)
0 10 20 30 40 50 60 70 80 90 100
Temps (min)
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
[A]∞
[A]0
0.15 min- 10.10 min- 1
0.07 min- 1
0.035 min- 1
ko b s = k1'
[B]0 = 0.033 M[B]0 = 0.065 M[B]0 = 0.098 M[B]0 = 0.145 M
Time 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16
[B]0 (M)
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
k obs
(min
-1)
pente = k2 = (1.02 ± 0.02) M- 1min- 1slope
78
CHM 8304
Kinetic analyses 40
Third order
• reactions that take place in one termolecular step are rare in the gas phase and do not exist in solution – the entropic barrier associated with the simultaneous collision of three
molecules is too high
A B P+k3
C+
Vitesse = v = = k3[A][B][C]d[P]dt
Rate
79
Third order, revisited
• however, a reaction that takes place in two consecutive bimolecular steps (where the second step is rate-limiting) would have a third order rate law!
A + B AB
C P
k1
+ABk2
Vitesse = v =d[P]dt
= k3[A][B][C]Rate
80
CHM 8304
Kinetic analyses 41
Zeroth order
• in the presence of a catalyst (organo-metallic or enzyme, for example) and a large excess of reactant, the rate of a reaction can appear to be constant
Vitesse = v = = kd[P]dt
= - d[A]dt
A Pk
catalyseur+ catalyseur+
− =∫ ∫d dtt
[A] kA
A
00[A]0 - [A] = k t [A] = -k t + [A]0 linear relation
[A]0 - ½ [A]0 = k t½
t½ 0[A]2 k
= half-life
catalyst catalyst
Rate
81
Zeroth order
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
pente = -k
not realistic that rate would be constant all the way to end of reaction...
Time
slope
82
CHM 8304
Kinetic analyses 42
Summary of observed parameters
Reaction order
Rate law Explicit equation Linear equation Half-life
zero d[P] kdt
= [A] = -k t + [A]0 [A]0 - [A] = k t t½ 0[A]2 k
=
first d[P] k [A]1dt= [A] = [A]0
k1e− t ln t[A][A]
k01= t ln
½1
2k
=
second d[P] k [A]22
dt= [A]
k[A]2
0
=+
11t
1 1[A] [A]
k0
2− = t t½2 0
1k [A]
=
d[P] k [A][B]2dt=
complex 1[B] [A]
[A] ([B] [P])[B] ([A] [P])
k0 0
0 0
0 02−
−
−
⎛
⎝⎜
⎞
⎠⎟ =ln t
complex
linear regression
non-linear regression
83
Exercise A: Data fitting • use conc vs time data to determine order of reaction and its rate
constant
0.0
20.0
40.0
60.0
80.0
100.0
0 10 20 30 40 50 60 70
Time (min)
[A] (
mM
)
Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.000 10 55.6 0.59 0.008 20 38.5 0.96 0.016 30 29.4 1.22 0.024 40 23.8 1.44 0.032 50 20.0 1.61 0.040 60 17.2 1.76 0.048
84
CHM 8304
Kinetic analyses 43
0.00
0.50
1.00
1.50
2.00
2.50
0 10 20 30 40 50 60 70
Time (min)
ln[A
]0/[A
] y = 0.0008x
0.000
0.010
0.020
0.030
0.040
0.050
0.060
0 10 20 30 40 50 60 70
Time (min)
1/[A
] - 1
[A]0
Exercise A: Data fitting • use conc vs time data to determine order of reaction and its rate
constant
linear non-linear
second order; kobs = 0.0008 mM-1min-1
85
Exercise B: Data fitting • use conc vs time data to determine order of reaction and its rate
constant
0.0
20.0
40.0
60.0
80.0
100.0
0 10 20 30 40 50 60 70
Time (min)
[A] (
mM
)
Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.00 10 60.7 0.50 0.006 20 36.8 1.00 0.017 30 22.3 1.50 0.035 40 13.5 2.00 0.064 50 8.2 2.50 0.112 60 5.0 3.00 0.190
86
CHM 8304
Kinetic analyses 44
Exercise B: Data fitting • use conc vs time data to determine order of reaction and its rate
constant
y = 0.05x
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50 60 70
Time (min)
ln ([
A]0/
[A]) y = 0.003x - 0.0283
-0.05
0
0.05
0.1
0.15
0.2
0.25
0 10 20 30 40 50 60 70
Time (min)
1/[A
] - 1
/[A]0
linear non-linear
first order; kobs = 0.05 min-1
87
Outline: Complex reactions
• based on A&D section 7.5 – consecutive first order reactions – steady state – changes in kinetic order – saturation kinetics – rapid pre-equilibrium
88
CHM 8304
Kinetic analyses 45
First order (consecutive)
Ak1
B Pk2
= k2 [B]Vitesse = v = d[P]dt ≠ -
d[A]dt
− =d[A] k [A]1dt
[A] = [A]0k1e− t
d[B] k [A] k [B]1 2dt= −
d t[B] k [A] k [B]1 0k
21
dte= −− d t[B] k [B] k [A]2 1 0
k1
dte+ = −
solved by using the technique of partial derivatives
[B] k [A](k k )
( )1
2 1
k k=−
−− −0 1 2e et t
[P] [A] [A] k [A](k k )
( )k 1
2 1
k k= − −−
−− − −0 0
01 1 2e e et t t [P] [A]k
(k k )( )k 1
2 1
k k= − −−
−⎛
⎝⎜
⎞
⎠⎟− − −
0 1 1 1 2e e et t t
Rate
89
Induction period
• in consecutive reactions, there is an induction period in the production of P, during which the concentration of B increases to its maximum before decreasing
A à B à P
• the length of this period and the maximal concentration of B varies as a function of the relative values k1 and k2 – consider three representative cases :
• k1 = k2 • k2 < k1
• k2 > k1
90
CHM 8304
Kinetic analyses 46
Consecutive reactions, k1 = k2
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
[A]0
[B]0, [P]0
[P]∞
[A]∞ , [B]∞
[B]max
pÈriode d'induction
k2 ≈ k1
0 10 20 30 40 50 60 70 80 90 100
Temps
-1
0
1
2
3
4
5
ln([P
] ∞ -
[P] 0)
- ln
([P] ∞
- [P
])
pente = kobs = k2 ≈ k1
k2 ≈ k1
pÈriode d'induction
induction period
induction period slope
Time
Time
91
Consecutive reactions, k2 = 0.2 × k1
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
[A]0
[B]max
[A]∞
[P]
[B]
k2 = 0.2 ◊ k1
0 10 20 30 40 50 60 70 80 90 100
Temps
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
ln([
P] ∞
- [
P] 0
) -
ln([
P] ∞
- [
P])
pÈriode d'induction
pente = kobs = k2
Time
induction period slope
Time
92
CHM 8304
Kinetic analyses 47
0 10 20 30 40 50 60 70 80 90 100
Temps
0
10
20
30
40
50
60
70
80
90
100
% [A
] 0
[A]0 [P]∞
[A]∞ , [B]∞
[B]max
k2 = 5 ◊ k1
0 10 20 30 40 50 60 70 80 90 100
Temps
-1
0
1
2
3
4
5
6
7
ln([P
] ∞ -
[P] 0)
- ln
([P] ∞
- [P
])
pÈriode d'induction
pente = kobs = k1
k2 = 5 ◊ k1
Consecutive reactions, k2 = 5 × k1
Time
induction period
slope
Time
93
Steady state • often multi-step reactions involve the formation of a reactive intermediate
that does not accumulate but reacts as rapidly as it is formed • the concentration of this intermediate can be treated as though it is
constant • this is called the steady state approximation (SSA)
94
CHM 8304
Kinetic analyses 48
Steady State Approximation
• consider a typical example (in bio-org and organometallic chem) of a two step reaction: – kinetic scheme:
– rate law:
– SSA :
• expression of [I] :
– rate equation:
A I Pk1
k-1
k2[B]
[ ] [ ][ ]BItP
2kdd
=
[ ] [ ] [ ] [ ][ ] 0BIIAtI
211 =−−= − kkkdd
[ ] [ ][ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BAI21
1
kkk
[ ] [ ][ ][ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BBA
tP
21
21
kkkk
dd first order in A;
less than first order in B
95
Steady State Approximation
• consider another example of a two-step reaction: – kinetic scheme:
– rate law:
– SSA:
• expression of [I] :
– rate equation:
[ ] [ ]ItP
2kdd
=
[ ] [ ][ ] [ ] [ ] 0IIBAtI
211 =−−= − kkkdd
[ ] [ ][ ]⎟⎟⎠
⎞⎜⎜⎝
⎛
+=
− 21
1 BAIkk
k
[ ] [ ][ ] [ ][ ]BABAtP
obs21
21 kkk
kkdd
=⎟⎟⎠
⎞⎜⎜⎝
⎛
+=
−
A + B I Pk1
k-1
k2
kinetically indistinguishable from the mechanism with no intermediate!
96
CHM 8304
Kinetic analyses 49
Steady State Approximation
• consider a third example of a two-step reaction: – kinetic scheme:
– rate law:
– SSA:
• expression of [I] :
– rate equation:
[ ] [ ][ ]BItP
22 kdd
=
[ ] [ ] [ ][ ] [ ][ ] 0BIPIAtI
2111 =−−= − kkkdd
[ ] [ ][ ] [ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BPAI
211
1
kkk
[ ] [ ][ ][ ] [ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BPBA
tP
211
212
kkkk
dd
A I + P1k1
k -1k2[B]
P2
first order in A, less than first order in B; slowed by P1
97
SSA Rate equations
• useful generalisations: 1. the numerator is the product of the rate constants and concentrations
necessary to form the product; the denominator is the sum of the rates of the different reaction pathways of the intermediate
2. terms involving concentrations can be controlled by varying reaction conditions
3. reaction conditions can be modified to make one term in the denominator much larger than another, thereby simplifying the equation as zero order in a given reactant
98
CHM 8304
Kinetic analyses 50
Change of reaction order
• by adding an exces of a reactant or a product, the order of a rate law can be modified, thereby verifying the rate law equation
• for example, consider the preceding equation:
– in the case where k-1 >> k2 , in the presence of excess B and (added) P1, the equation can be simplified as follows:
• now it is first order in A and in B
[ ] [ ][ ][ ] [ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BPBA
tP
211
212
kkkk
dd
[ ] [ ][ ][ ] ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− 11
212
PBA
tP
kkk
dd
99
Change of reaction order
• however, if one considers the same equation:
– but reaction conditions are modified such that k-1[P1] << k2[B] , the equation is simplified very differently:
• now the equation is only first order in A
• in this way a kinetic equation can be tested, by modifying reaction conditions
[ ] [ ][ ][ ] [ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BPBA
tP
211
212
kkkk
dd
[ ] [ ][ ][ ] ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
BBA
tP
2
212
kkk
dd
[ ] [ ]AtP
12 kdd
=
100
CHM 8304
Kinetic analyses 51
Saturation kinetics
• on variation of the concentration of reactants, the order of a reactant may change from first to zeroth order – the observed rate becomes “saturated” with respect to a reactant – e.g., for the scheme
having the rate law
one observes:
A I Pk1
k-1
k2[B]
[ ] [ ][ ][ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− BBA
tP
21
21
kkkk
dd
v
[B]0
vmax = k1[A]
first order in B
~zero order in B
101
Example of saturation kinetics
• for a SN1 reaction, we are taught that the rate does not depend on [Nuc], but this is obviously false at very low [Nuc], where vobsà0 all the same
• in reality, for
one can show that • but normally k2 >> k-1 and [-CN] >> [Br-]
so the rate law becomes:
• i.e., it is typically already saturated with respect to [-CN]
Brk1
k -1+ Br-
k2 [-CN] CN
[ ] [ ][ ][ ] [ ]⎟
⎟⎠
⎞⎜⎜⎝
⎛
+=
− CNBrCNR
tP
21
21
kkkk
dd
[ ] [ ][ ][ ]
[ ]RCNCNR
tP
12
21 kkkk
dd
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
102
CHM 8304
Kinetic analyses 52
Rapid pre-equilibrium
• in the case where a reactant is in rapid equilibrium before its reaction, one can replace its concentration by that given by the equilibrium constant
• for example, for the protonated alcohol is always in rapid eq’m with the alcohol and therefore
• normally, k2[I-] >> k-1[H2O] and therefore
OHk1
k -1+ H2O
k2 [-I] IH
OHKeq[H+]
[ ] [ ][ ][ ][ ] [ ] ⎟⎟
⎠
⎞⎜⎜⎝
⎛
+
+=
− IOHIBuOHH
tP
221
21
kktKkk
dd eq
[ ] [ ][ ][ ][ ]
[ ][ ]BuOHHI
IBuOHHtP
12
21 tKkktKkk
dd
eqeq +=⎟⎟
⎠
⎞⎜⎜⎝
⎛ += first order in tBuOH,
zero order in I-, pH dependent
103
Outline: Multiple reaction co-ordinates
• based on A&D section 7.8 – variation of activated complexes – More-O’Ferrall-Jencks diagrams – vibrational state effects
104
CHM 8304
Kinetic analyses 53
Nucleophilic substitution reactions • SN1 and SN2 reactions both involve the cleavage of the C-LG bond and
the formation of the C-Nuc bond
• if these events are assigned to the x and y axes of an energy surface, one can visualise how the structure of a substrate will determine which of the two mechanisms is favoured:
105
More-O’Ferrall-Jencks (MOFJ) diagram • projection of an energy surface on a two-dimensional plot
– (e.g. Figure 7.21 A&D) :
note the displacement of the TS from an SN2 reaction to an SN1
106
CHM 8304
Kinetic analyses 54
Displacement of a transition state • on a MOFJ diagram, one raises or lowers the corners of the plot, as a
function of the change of structure of a species and then considers how the transition state would be displaced, according to two effects:
‡
1. “Hammond effect”: if the corner of the reactants or products is lowered, the transition state is displaced along the diagonal of the pathway, towards the opposite corner, according to the Hammond postulate
2. “orthogonal effect” : if an “off-diagonal” corner is lowered, the transition state is displaced towards the lowered corner
‡
‡ ‡
R
P
I1
I2 R
P
I1
I2 107
More-O’Ferrall-Jencks Diagram
‡ ‡
Et GP + Nuc Et Nuc + GPLG LG
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
GP
Nucδ−
δ−Me
H
H
LG
symmetric
108
CHM 8304
Kinetic analyses 55
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
More-O’Ferrall-Jencks Diagram
‡
Et GP + Nuc Et Nuc + GPbetter
‡ ‡
LG LG
GP
Nucδ−
δ−Me
H
Hδ−
LG
109
“Hammond effect”
“orthogonal effect”
less C-LG cleavage
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
More-O’Ferrall-Jencks Diagram
‡
worse
‡ ‡
Et GP + Nuc Et Nuc + GPLG LG
GP
Nucδ−
δ−
MeH
Hδ+
LG
110
more C-LG cleavage
CHM 8304
Kinetic analyses 56
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
More-O’Ferrall-Jencks Diagram
‡
better
‡
‡
Et GP + Nuc Et Nuc + GPLG LG
GP
Nucδ−
δ−Me
H
Hδ+
LG
less C-Nuc formation
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
More-O’Ferrall-Jencks Diagram
‡
better
‡
‡
worse Et GP + Nuc Et Nuc + GPLG LG
GP
Nucδ−
δ−
MeH
Hδ+
LG
112
dissociative
CHM 8304
Kinetic analyses 57
Cleavage of C-LG
Form
atio
n of
C-N
uc
GP + Nuc
Nuc + GP
+ Nuc+ GPEt
H3C CH2
Nuc
GP
LG
LG LG
LG
More-O’Ferrall-Jencks Diagram
‡
worse
‡
worse Et GP + Nuc Et Nuc + GP
‡
LG LG
GP
Nucδ−
δ−
MeH
H
LG resembles products!
Vibrational effects
• certain vibrational modes of certain bonds may assist in the formation of a given activated complex – known as productive vibrations
• on the other hand, other vibrational modes may impede attainment of the transition state – known as non-productive vibrations
• the impact of these vibrational effects depends on the reaction and the nature of its transition state(s)
• the shape of the energy surface of the reaction reflects this dependence and helps to visualise the entropic effect of changes of the degrees of freedom on passing from a reactant to the transition state
114
CHM 8304
Kinetic analyses 58
Shape of an energy surface • the width of a “valley” on a surface is related to the
number of degrees of freedom, and therefore entropy, of a given molecule
1. when the col is as narrow as the initial valley, the activated complex has the same shape (and degrees of freedom) as the reactant and ΔS‡ ≈ 0
2. when the col is wider than the initial valley, there are more degrees of freedom at the transition state, and ΔS‡ > 0
3. when the col is narrower than the initial valley, there are fewer degrees of freedom in the activated complex, and ΔS‡ < 0
1.
2.
3. Figure 7.23, A&D 115
Temperature and energy surfaces
• at higher temperature, molecules are more excited and can navigate more easily through broad cols than through narrow cols
• this analogy helps us understand how entropy contributes to the free energy of activation at higher temperatures, according to the following equation:
ΔG‡ = ΔH‡ - TΔS‡
116
CHM 8304
Kinetic analyses 59
Summary: Kinetic approach to mechanisms
• kinetic measurements provide rate laws – ‘molecularity’ of a reaction
• rate laws limit what mechanisms are consistent with reaction order – several hypothetical mechanisms may be proposed
• detailed studies (of substituent effects, etc) are then necessary in order to eliminate all mechanisms – except one! – one mechanism is retained that is consistent with all data – in this way, the scientific method is used to refute inconsistent
mechanisms (and support consistent mechanisms)
117